# Lagrange classical relation

I have been studying theoretical mechanics and just now I came cross a formula called "Lagrange classical relation", that is, if we let $q_1$, $q_2$,$\cdot $$\cdot$$\cdot$, $q _ m$, $t$ be the $generalized$ $coordinates$ of $x$, then we have $${\partial\dot{ x}\over\partial \dot{q _ k}}={\partial x\over\partial q_ k}$$ My understanding is, given a differentiable map $x$: $\Bbb {R}$$\to$$\Bbb {R}$, $t$ $\mapsto$ $x(t)$,
and another differentiable map $\phi$: $\Bbb {R}^{m+1}$$\to$$\Bbb {R}$, ($q_1$, $q_2$,$\cdot $$\cdot$$\cdot$,$t$) $\mapsto$$\phi( q_1, q_2,\cdot$$\cdot $$\cdot ,t). And we have another map g: \Bbb {R}$$\to$ $\Bbb {R}^{m+1}$, $t$$\mapsto$$($$q_1(t), q_2(t),\cdot$$\cdot $$\cdot ,t$$)$, which satisfys $\phi$ $\circ$ $g$ $=$ $x$.
And I can calculate that
$$\dot{x}(t)=\sum_{i=1}^m[({\partial \phi\over \partial q _ k})\circ g](t)\cdot\dot{q_i}(t)+({\partial \phi\over \partial {t}})\circ g(t), or$$$$\dot{x}=\sum_{i=1}^m[({\partial \phi\over \partial {q _ k}})\circ g]\cdot\dot{q_i}+({\partial \phi\over \partial {t}})\circ g,$$ ,in the second formular I am using the operations between mappings.
But I think that $\dot{x}$ is a map $\Bbb {R}$$\to$$\Bbb {R}$, so I cannot understand the meaning of mathematical symbol$${\partial\dot{ x}\over \partial \dot{q _ k}}$$,since I don't think we can talk about the partial derivatives of $\dot{x}$.
Question: How should I understand the partial derivative symbol $${\partial\dot{ x}\over \partial \dot{q _ k}}?$$

In Lagrangian mechanics we use independent generalized coordinates and express every cartesian coordinate using these generalized coordinates:

$$\begin{cases} x_{1} \equiv x_{1}(q_{1}(t),.., q_{n}(t), t) \\ ...\\ x_{N} \equiv x_{N}(q_{1}(t),.., q_{n}(t), t) \end{cases}$$

As you can see every $x_{n}$ is a composite function, so when we want the total derivative of $x_{n}$ we simply apply the chain rule:

$$\frac{dx}{dt} = \sum_{l=1}^{k}\frac{\partial x}{\partial q_{l}}(q_{1},...,q_{n}, t) \dot{q_l}(t) + \frac{\partial x}{\partial t}(q_{1},...,q_{n}, t)$$

So the time derivative of every cartesian coordinate is a function of the generalized coordinates and their time derivatives and time. $\dot{x}: (q_1,..., q_n, \dot{q_1},... \dot{q_n}, t) \rightarrow \dot{x}$

We can easily take the partial derivative of this function $\dot{x}$ with respect to the time derivative of a specific generalized coordinate:

$$\frac{\partial}{\partial \dot{q_k}} \left ( \dot{x}\right ) = \frac{\partial x}{\partial q_k}$$

Well, I'd like to give my personal idea:
Since we have $$\dot{x}(t)=\sum_{i=1}^m[({\partial \phi\over \partial q _ k})\circ g](t)\cdot\dot{q_i}(t)+({\partial \phi\over \partial {t}})\circ g(t), \forall t\in \mathbb{R},$$ Then if we define the following maps: $$\mu:\mathbb{R}^{2m+1}\to \mathbb{R},(y_1,y_2,\cdots,y_{2m+1})\mapsto \sum_{i=1}^my_{m+i}\cdot y_i\ +y_{2m+1},and$$$$\Pi:\mathbb{R}\to \mathbb{R}^{2m+1}, t\mapsto\left(\dot{q_1}(t),\cdots,\dot{q_m}(t),[({\partial \phi\over \partial q _ 1})\circ g](t),\cdots,[({\partial \phi\over \partial q _ m})\circ g](t),[({\partial \phi\over \partial {t}})\circ g](t)\right) ,$$ then we have $$\dot{x}(t)=(\mu\circ\Pi)(t),\forall t\in \mathbb{R}$$ thus $\dot{x}=\mu\circ\Pi.$ And we can see $${\partial \mu\over \partial y_ k}(z_1,z_2,\cdots,z_{2m+1})=z_{m+k},\forall 1\leqslant k \leqslant m,$$ thus$$\left({\partial \mu\over \partial y_ k}\circ\Pi\right)(t)=[({\partial \phi\over \partial q _ k})\circ g](t),\forall 1\leqslant k\leqslant m.$$ So when we say $${\partial\dot{ x}\over \partial \dot{q _ k}}={\partial x\over\partial q_ k},$$our actual meaning is that $$\left({\partial \mu\over \partial y_ k}\circ\Pi\right)=[({\partial \phi\over \partial q _ k})\circ g].$$ We also have $\dot{x}=\mu\circ\Pi ,and \ x=\phi\circ g,$ which is interesting.

But I don't this explanation is convincing, and I do think that the symbol $${\partial\dot{x}\over \partial \dot{q _ k}}$$is actually very misleading.

For how I see it, the procedure of replacing your $$x_i,\dot x_i$$ coordinates with a new set of generalized $$q_i,\dot q_i$$, can be seen as the choice of the unique transformation from x to q that would deliver the right motion, and this may be thought as the extremization of the functional

$$\int_{t_1}^{t_2} x[q,\dot q;t] dt$$

In the family of trajectories that keep $$q_i(t_1)$$ and $$q_i(t_2)$$ fixed.

Then, Eulero-Lagrange equations analogous to the une used in Lagrangian mechanics would yield

$$\frac d{dt}\frac{\partial x}{\partial \dot q}= \frac{\partial x}{\partial q}$$

That, exchanging the derivation signs in the lhs, becomes:

$$\frac{\partial \dot x}{\partial \dot q}= \frac{\partial x}{\partial q}$$

I might be totally wrong but it looks nice;